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Posted: 10/2/2013 8:09:48 PM EDT
Can someone explain to me how to answer the the question boxed in below given the information in the graph and instructions?  Thanks for any help.

Edit:  Damn, tiny pic shrunk my picture.  The upper curve is the line 6x + y = 12.  The bottom curve is y = x^2 - 4.  I need to approximate the area of the vertical strip with respect to x.  I was able to figure out the area between the curves, but I cheated by using formulas for the figures instead of the definite integral representing the area.

Link Posted: 10/2/2013 8:20:18 PM EDT
[#1]
What is the square root of 87?

ETA- sorry been drinking, thought I was in GD.  I took Calc, Diff Eqs, etc over 20 years ago, I am of no use.
Link Posted: 10/2/2013 10:25:31 PM EDT
[#2]
Okay, I figured it out, but I need someone to help explain to me how I got to where I did.  The answer I got for the approximation of the area of the vertical slice with respect to x is [ ( 12 - 6x ) - ( x^2 - 4 ) ] delta(x).  Originally I kept putting a positive sign between the functions but then it occurred to me that Integral ( x^2 -4 ) dx is negative over the range of integration.  My line of reasoning is that what the question was asking was for the area, which must be positive, thus I needed the absolute value of the integral, which means I needed to subtract the negative to make it positive.  Is that right, or am I just a blind squirrel that managed to find a nut?
Link Posted: 10/3/2013 2:02:47 AM EDT
[#3]
Ignor the 2 equations they gave you. Pretend the line and curve are above the x axis. You are subtracting the area of the bottom curve  to the x axis from the area of the top line to the x axis. Always check your answers with exact values.
Link Posted: 10/3/2013 4:05:19 AM EDT
[#4]
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Quoted:
Ignor the 2 equations they gave you. Pretend the line and curve are above the x axis. You are subtracting the area of the bottom curve  to the x axis from the area of the top line to the x axis. Always check your answers with exact values.
View Quote


Link Posted: 10/3/2013 9:31:10 AM EDT
[#5]
Discussion ForumsJump to Quoted PostQuote History
Quoted:
Okay, I figured it out, but I need someone to help explain to me how I got to where I did.  The answer I got for the approximation of the area of the vertical slice with respect to x is [ ( 12 - 6x ) - ( x^2 - 4 ) ] delta(x).  Originally I kept putting a positive sign between the functions but then it occurred to me that Integral ( x^2 -4 ) dx is negative over the range of integration.  My line of reasoning is that what the question was asking was for the area, which must be positive, thus I needed the absolute value of the integral, which means I needed to subtract the negative to make it positive.  Is that right, or am I just a blind squirrel that managed to find a nut?
View Quote


When integrating, you already know that you are finding the area under the curve. Area ABOVE the x-axis is positive, and BELOW the x-axis is negative.

What you end up finding is the NET area, with an integral. You're still adding the two areas together, except one is negative, which of course leads to subtraction.

Area above x + (- Area below x)

Area under the curve is a good visual representation of integration, but the term, "area," comes with a limitation that integration doesn't have. By common sense, "area," can't be negative, right? An integral can.

Try integrating x^2 - 6 from 0 to 1, just to convince yourself that integrals can be negative, as well

You're on the right path. Area above, positive, area below, negative.
Link Posted: 10/3/2013 3:59:21 PM EDT
[#6]
Discussion ForumsJump to Quoted PostQuote History
Quoted:


When integrating, you already know that you are finding the area under the curve. Area ABOVE the x-axis is positive, and BELOW the x-axis is negative.

What you end up finding is the NET area, with an integral. You're still adding the two areas together, except one is negative, which of course leads to subtraction.

Area above x + (- Area below x)

Area under the curve is a good visual representation of integration, but the term, "area," comes with a limitation that integration doesn't have. By common sense, "area," can't be negative, right? An integral can.

Try integrating x^2 - 6 from 0 to 1, just to convince yourself that integrals can be negative, as well



You're on the right path. Area above, positive, area below, negative.
View Quote View All Quotes
View All Quotes
Discussion ForumsJump to Quoted PostQuote History
Quoted:
Quoted:
Okay, I figured it out, but I need someone to help explain to me how I got to where I did.  The answer I got for the approximation of the area of the vertical slice with respect to x is [ ( 12 - 6x ) - ( x^2 - 4 ) ] delta(x).  Originally I kept putting a positive sign between the functions but then it occurred to me that Integral ( x^2 -4 ) dx is negative over the range of integration.  My line of reasoning is that what the question was asking was for the area, which must be positive, thus I needed the absolute value of the integral, which means I needed to subtract the negative to make it positive.  Is that right, or am I just a blind squirrel that managed to find a nut?


When integrating, you already know that you are finding the area under the curve. Area ABOVE the x-axis is positive, and BELOW the x-axis is negative.

What you end up finding is the NET area, with an integral. You're still adding the two areas together, except one is negative, which of course leads to subtraction.

Area above x + (- Area below x)

Area under the curve is a good visual representation of integration, but the term, "area," comes with a limitation that integration doesn't have. By common sense, "area," can't be negative, right? An integral can.

Try integrating x^2 - 6 from 0 to 1, just to convince yourself that integrals can be negative, as well



You're on the right path. Area above, positive, area below, negative.



this
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